Let $$a_1, a_2, a_3, \ldots$$ be a G.P. of increasing positive numbers. Let the sum of its 6$$^{th}$$ and 8$$^{th}$$ terms be 2 and the product of its 3$$^{rd}$$ and 5$$^{th}$$ terms be $$\frac{1}{9}$$. Then $$6a_2 + a_4a_4 + a_6$$ is equal to

Given:

Let the first term of the G.P. be $$a > 0$$ and the common ratio be $$r > 1$$ (since the terms are increasing positive numbers).

1. Sum of $$6^{\text{th}}$ and $8^{\text{th}}$$ terms:

   $$a_6 + a_8 = 2 \implies a r^5 + a r^7 = 2 \implies a r^5 (1 + r^2) = 2 \quad \text{--- (i)}$$

2. Product of $3^{\text{rd}}$ and $5^{\text{th}}$ terms:

   $$a_3 \cdot a_5 = \frac{1}{9} \implies (a r^2)(a r^4) = \frac{1}{9} \implies a^2 r^6 = \frac{1}{9}$$

   Taking the square root on both sides ($$a, r > 0$$):

   $$a r^3 = \frac{1}{3} \quad \text{--- (ii)}$$

Finding $$r$$ and $$a$$:

Rewrite equation (i) by factoring out $$a r^3$$:

$$(a r^3) \cdot r^2 (1 + r^2) = 2$$

Substitute $$a r^3 = \frac{1}{3}$$ from equation (ii):

$$\frac{1}{3} r^2 (1 + r^2) = 2 \implies r^2 (1 + r^2) = 6$$
$$r^4 + r^2 - 6 = 0$$$$(r^2 + 3)(r^2 - 2) = 0$$

Since $$r^2 > 0$$ for real numbers, $$r^2 = 2 \implies r = \sqrt{2}$$.

Substitute $$r = \sqrt{2}$$ back into equation (ii) to find $$a$$:

$$a r^3 = \frac{1}{3} \implies a(\sqrt{2})^3 = \frac{1}{3} \implies a(2\sqrt{2}) = \frac{1}{3} \implies a = \frac{1}{6\sqrt{2}}$$

Evaluating $$6(a_2 + a_4)(a_4 + a_6)$$:

Express the terms using $$a$$ and $$r$$:

$$a_2 + a_4 = a r + a r^3 = a r (1 + r^2)$$
$$a_4 + a_6 = a r^3 + a r^5 = a r^3 (1 + r^2)$$

Multiplying both expressions:

$$6(a_2 + a_4)(a_4 + a_6) = 6 \cdot [a r (1 + r^2)] \cdot [a r^3 (1 + r^2)] = 6 a^2 r^4 (1 + r^2)^2$$

Substitute $$a = \frac{1}{6\sqrt{2}} \implies a^2 = \frac{1}{72}$$, $$r^2 = 2 \implies r^4 = 4$$, and $$(1+r^2) = 3$$:

$$= 6 \times \left(\frac{1}{72}\right) \times 4 \times (3)^2$$
$$= \frac{6 \times 4 \times 9}{72} = \frac{216}{72} = 3$$

Correct Answer: 3